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Work done by force (Vector notation) 
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#1
Apr311, 11:22 AM

P: 42

1. The problem statement, all variables and given/known data
How much work is done by a force= (2x)i N + (3)j N with x in meters, that moves a particle from a position (initial) = (2m)i + (3m)j to a position (final) = (4m)i + (3m)j? Answer:  6 J 2. Relevant equations Work = integral of Force 3. The attempt at a solution W = [x^2] i + [3x] j W = [4^2  2^2] i + [3(3)  3(3)] j W = 12 i 18 j W magnitude = 21.6J I realize I am doing this wrong becasue the Magnitude will always be positive and the answer (from back of book) is negative. I tried this in a similar way where I took the difference between the two position vector and then placed it in the integral rd = (6) i + (6) j Got W = 36 i + 18 j still wrong. Help, pretty please? 


#2
Apr311, 12:20 PM

P: 1,402

To save typing all the i's and j's, I'll use the notation (2,3) = 2i + 3j. I'll also cut corners by leaving out the units.
Assuming the movement is in a straight line, we can parameterize the particle's path from the initial point (2,3) to the final point (4,3) by an equation such as [tex]\mathbf{r}(t) = (x,y) = (2,3)  t(6,6), \enspace 0 \leq t \leq 1.[/tex] (6,6) is the displacement vector from the initial point to the final point, that is, the difference between them. Then [itex]x=26t[/itex], and [tex]d\mathbf{r} = (dx,dy) = (6,6) dt.[/tex] So [tex]\int_{\mathbf{r}(0)}^{\mathbf{r}(1)} \mathbf{F}(\mathbf{r}) \cdot d\mathbf{r}[/tex] [tex]=\int_{(2,3)}^{(4,3)}(2x,3) \cdot (dx,dy) = \int_{0}^{1}(4272t) \; dt = 6[/tex] 


#3
Apr311, 01:16 PM

P: 42




#4
Apr311, 02:21 PM

P: 1,402

Work done by force (Vector notation)
Oopsh, sorry, I missed out a minus in front of the integral on the right. Are the following steps clear?
[tex]x = 2  6t[/tex] [tex](2x,3) = (2(26t),3) = (412t,3)[/tex] [tex](412t,3) \cdot (6,6) = 6(412t)  18 = 42 + 72t.[/itex] (6,6) comes from differentiating [itex]\mathbf{r}[/itex]. 


#5
Apr911, 05:07 PM

P: 42

My test is this Wednesday, but I have lots of other tests so sorry about the delay. A couple things: 1) where did time come from? The question never mentioned it only position. 2) I understand this: r = (6) i + (6) j r = (6,6) 3) where does this come from: x=26t (I think this is an extension of being confused about the time)? Why is integral not (x^2)i + (3x) j ??? Thank you again for your help and I am sorry if I am difficult. 


#6
Apr1011, 09:24 AM

P: 42

The reason time confuses me is that isn't POWER change in work over time? I thought the difference between Force and power is whether your taking the derivative of work with respect to time or with respect to position....?



#7
Apr1011, 01:10 PM

P: 42

Okay, I figured out what I did wrong. I had this part right:
W = [x^2] i + [3x] j W = [4^2  2^2] i + [3(3)  3(3)] j W = 12 i 18 j What I needed to do was subtract 18 from 12 W = 12  18 = 6 (answer!) I wasn't doing this becasue I thought they were sepearte becasue of the separate components. Wow, I can't believe I was so close for so long! I still don't completely understand why you just add i and j together to find work instead of finding the magnitude... but it works for many similar book problems so I know this is the right way to do it. Thanks for the help. :) 


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